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I am going through $IB$ further math geometry topic and just learned Ceva's theorem. Below is one of the questions in the exercise. I have thought for quite a while and cannot solve it. Can anyone give me some clues or hints, please? Thanks.

In the diagram, $BZ:ZC=2:1$ and $AR:RS:SZ=5:4:3$. Find the ratio in which $X$ divides $[AB]$.

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nonuser
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Fellow InstituteOfMathophile
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3 Answers3

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The proof of Ceva'sTheorem uses the fact that if $ A, B $ and $ C $ are three non-collinear points then any vector $ P $ can be expressed as $ P = xA + yB + zC $ where $ x+ y + z = 1 $. Also, a point on the line joining $ A $ and $ B $ is given by $ P = tA + (1-t)B $.

$ Z = \frac {2}{3} C + \frac {1}{3} B $

$ S = \frac {3}{4} Z + \frac {1}{4} A $ $ = \frac {1}{2} C +\frac {1}{4} B + \frac {1}{4} A $

Similarly, $ R = \frac {5}{18} C + \frac {5}{36} B + \frac {7}{12} A $

$ Y = tS + (1-t)B $ $ = \frac {t}{2} C + (\frac {t}{4} + 1-t)B + \frac {t}{4} A $

For $ Y $ to lie on $ AC $, $ t = \frac {4}{3} $ . Then $ Y = \frac {2}{3} C + \frac {1}{3} A $

$ X = kR + (1-k)Y $

This gives $ X $ in terms of $ A, B $ and $ C $.

Find $ k $ for which $ X $ lies on $ AB $ etc.

kieran
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  • I am learning Ceva's theorem in purely geometric way, not in vector form. Anyway your solution gives me an alternative way of thinking it. Especially the part that three nonlinear points can be represented in that way. I never know that and thus tried to prove it on my own. It was interesting. Thank you for your effort. – Fellow InstituteOfMathophile Jan 18 '18 at 04:19
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Here, for a start, we have triangle $AZC$ and transversal $Y-B-S$:

$$ {AY\over YC} \cdot {CB\over ZB}\cdot{ZS \over SA} =1$$ so $${AY\over YC} = 2$$

Then we have triangle $BCY$ and transversal $A-S-Z$:

$$ {AY\over CA}\cdot {CZ\over ZB}\cdot {BS \over SY} =1$$ so $${BS\over SY} = 3$$

Finally we have triangle $ABS$ and transversal $Y-R-X$. Can you finish?

nonuser
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Here is one more alternative solution with homotheties:

Let $\mathcal{H}_{T,k}$ denote a homothety with center at $T$ and a extension factor $k$.

We have a following fact (Theorem):

If $\mathcal{H}_{M,k_1}$ and $\mathcal{H}_{N,k_2}$ are homotheties then their compostion $\mathcal{H}_{M,k_1}\circ \mathcal{H}_{N,k_2}$ is again some homothety $\mathcal{H}_{K,k}$ with $k=k_1k_2$ (if $k\ne 1$) and it center $K$ lies on a line $MN$.

Since we have: \begin{align} \mathcal{H}_{A, {4\over 3}}: & \;S \longmapsto Z\\ \mathcal{H}_{C,{3}}: &\; Z \longmapsto B\\ \end{align} we see that $Y$ is a center of homothety which takes $S$ to $B$ with ratio $k=4$.

Now again \begin{align} \mathcal{H}_{R, {-4\over 5}}: & \;A \longmapsto S\\ \mathcal{H}_{Y,{4}}: &\; S \longmapsto B\\ \end{align} we see that $X$ is a center of homothety which takes $A$ to $B$ with ratio $k={-16\over 5}$, so $\boxed{AX:XB = 5:16}$.

nonuser
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